CBSE Class 12-science Answered

Anions ccp (fcc). Radius Na⁺ = 1.02Å, radius Cl^- = 1.81Å; radius ratio = 0.563.

Therefore Na octahedral.

1 octahedral / anion therefore 100% octahedral sites are filled.

Coordination # Na = 6; coordination # Cl = 6. 
For an fcc lattice there are 4 lattice points per cell, the motif in this case is a Na cation and a Cl anion.  Therefore the cell contents are 4 Na cations + 4 Cl anions.

Lattice parameter and Density calculations.

The cation and its nearest neighbor anions are, by definition in contact.
Therefore the cell edge  = 2r(Na⁺) + 2r(Cl^-) = 2(1.02) + 2(1.81) = 5.66Å

The Coulombic interactions involve all the surrounding ions:

  6 Cl^-    at a distance    d

12 Na⁺    at a distance   d√2

  8 Cl^-   at a distance   d√3

  6 Na⁺    at a distance   d√4 = 2d

24 Cl^-    at a distance   d√5

                                                      and so on…

Thus we have

     U_C  =  Z⁺Z^-e² ( 6 – 12/√2 + 8/√3 – 6/√4 + 24/√5 - ...)

                                                               4pe_od

The infinite series can be summed, and Madelung’s constant for the NaCl structure type  M_NaCl = 1.748.